posted by hammie .
I'm suppose to calculate the weight percent per volume (grams pre 100cm3) of ethanoic acid in a commercial sample of vineger from the following: (a) 0.4321g of pure momoprotic acid( formula weight 204.2 amu) required 23.45cm3 of naoh for neutralizalion.
You need to know the molarity of the NaOH.
The 204.2 sounds like you standardized the NaOH by using 0.4321 g of pure potassium hydrogen phthalate.
moles KHP = 0.4321/204.22 = ?
moles NaOH = moles KHP
M NaOH = moles NaOH/0.02435
You can work out the M NaOH.
Now all we need to know is the volume of the vinegar sample you used.
M NaOH x L NaOH = moles NaOH used for the sample.
moles NaOH for the sample = moles acid in the vinegar.
grams acid in the vinegar = moles x molar mass (60).
All of that gives you grams acetic acid/volume used. Convert that to g/100 cc and you have it.
Post your work if you get stuck and PLEASE type ALL of the problem.